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Let $C_k$ be the $k$-simplex. I know that $$\int_{C_k} \prod_{i=1}^k x_i^{\alpha_i-1} dx_i = \frac{\prod_{i=1}^k \Gamma(\alpha_i)}{\Gamma\left(\sum_{i=1}^k \alpha_i\right)} \equiv B(\alpha_1,\ldots,\alpha_k)$$ where $B$ is called the multinomial beta function.

What is the following integral? $$\int_{C_k} \prod_{i=1}^k dx_i\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}$$

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  • $\begingroup$ if i'm not totally wrong you just have to replace $\alpha_i$ by $\beta_i=2\alpha_{ii}+\sum_{j>i}^k (\alpha_{ij})-(k+1)$. You need to prove by induction that the inner product equals $x_i^{\beta_i}$ . $\endgroup$ – tired Dec 18 '14 at 18:28
  • $\begingroup$ would you mind expanding that? I have no clue as to where your $\beta_i$ comes from. $\endgroup$ – yannick Dec 18 '14 at 18:48
  • $\begingroup$ What happens if you expand this product? $\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}$ Maybe it's best to start with two easy examples, $k=2,3$. Please note also, that my former comment contains a litte mistake. You should replace $\alpha_i-1$ by $\beta_i$ $\endgroup$ – tired Dec 18 '14 at 18:55
  • $\begingroup$ of course! I must have been tired. I will write an answer tomorrow, but please, feel free to do it today :) $\endgroup$ – yannick Dec 18 '14 at 19:01
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Let's suppose that we are aware of the integral $$\int_{C_k} \prod_{i=1}^k x_i^{\alpha_i-1} dx_i = \frac{\prod_{i=1}^k \Gamma(\alpha_i)}{\Gamma\left(\sum_{i=1}^k \alpha_i\right)} $$ $(\textbf{1})$

(See here for a proof).

It turns out that only thing we need is to rewrite properly the inner one of the two products:

$$ P(k)=\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1} \text{ =?} $$ Let's try the easy example $k=3$, (assuming the $\alpha_{ij}$ are symmetric)we get $$ (x_1x_1)^{\alpha_{11}-1}(x_1x_2)^{\alpha_{12}-1}(x_1x_3)^{\alpha_{13}-1}(x_2x_2)^{\alpha_{22}-1}(x_2x_3)^{\alpha_{23}-1}(x_3x_3)^{\alpha_{33}-1} $$ which equals $$ x_1^{2\alpha_{11}+\alpha_{12}+\alpha_{13}-4}x_2^{2\alpha_{22}+\alpha_{12}+\alpha_{23}-4}x_3^{2\alpha_{33}+\alpha_{13}+\alpha_{23}-4} $$ or $$ P(3)=\prod_i^3 x_i^{2\alpha_{ii}+\sum_{j\neq i}^3\alpha_{ij}-4} $$ It is obvious (or can be shown by induction) that one can generalize this to arbitrary $k$ $$ P(k)=\prod_{i=1}^k x_i^{\alpha_{ii}-1+\sum_{j=1}^k(\alpha_{ij}-1)} $$ Plugging this expression into our integral (using $\prod_ib_i\prod_i a_i=\prod_i a_i b_i$) we see that it is given by $$\int_{C_k} \prod_{i=1}^k dx_i\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}=\int_{C_k} \prod_{i=1}^k dx_ix_i^{\alpha_{ii}-1+\sum_{j=1}^k(\alpha_{ij}-1)}$$

which allows us to just use $(\textbf{1})$, replacing $\alpha_i$ by $\alpha_{ii}+\sum_{j=1}^k(\alpha_{ij}-1)$

$$ \int_{C_k} \prod_{i=1}^k dx_i\prod_{j=i}^k (x_i x_j)^{\alpha_{ij}-1}=\frac{\prod_{i=1}^k \Gamma\left(\alpha_{ii}+\sum_{j=1}^k(\alpha_{ij}-1)\right)}{\Gamma\left(\sum_{i=1}^k \left[\alpha_{ii}+\sum_{j=1}^k(\alpha_{ij}-1)\right]\right)} $$

Edit: It appears that in my original answer, i assumed implicitly that the $a_{ij}$ are symmetric $\alpha_{ij}$=$\alpha_{ji}$. If this is not the case, one has to modify the answer as follows: Replace $\sum_{j\neq i}^k\alpha_{ij}$ by $\sum_{j\neq i}^k\left(\Theta(i-j)\alpha_{ij}+\Theta(j-i)\alpha_{ji}\right)$ where $\Theta(x)$ is Heaviside's step function.

Please note also that this approach is generalizable to integrals like $$ \int_{C_k} \prod_{i=1}^k dx_i\prod_{i_1, i_2, \dot .......,i_n=i}^k (x_{i_1} x_{i_2}...... x_{i_n})^{\alpha_{i_1 i_2...i_n}-1} $$

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  • $\begingroup$ Thanks for cleaning up the mess! I had to leave yesterday suddenly... :( $\endgroup$ – tired Dec 19 '14 at 10:38
  • $\begingroup$ Please see my edit... $\endgroup$ – tired Dec 19 '14 at 10:45

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